设f(x)=asin2x+bcos2x若f（x）小于等于f（π/6）的绝对值对一切x属于实数恒成立则f（x）的单调递增区

设f(x)=asin2x+bcos2x=根号（a²+b²）sin(2x+φ)
因为 f（x）小于等于f（π/6）的绝对值对一切x属于实数恒成立
所以 x=π/6时 |sin(2x+φ)|=1
2π/6+φ=kπ+π/2
所以 φ=kπ+π/6 k∈Z
若k=2n,n∈Z,f(x)=根号（a²+b²）sin(2x+π/6)
由2nπ-π/2≤2x+π/6≤2nπ+π/2
得nπ-π/3≤x≤nπ+π/6
单调递增区间为[nπ-π/3 ,nπ+π/6] n∈Z
若k=2n-1,n∈Z,f(x)=根号（a²+b²）sin(2x-5π/6)
由2nπ-π/2≤2x-5π/6≤2nπ+π/2
得nπ+π/6≤x≤nπ+2π/3
单调递增区间为[nπ+π/6 ,nπ+2π/3] n∈Z